Quantum String action

The variation of the field $B$ vanishes at the boundary thus the total derivative term vanishes. Now regarding the last term, the derivative should also act on the exponential. When you apply Leibniz rule you have:

$$\partial(e^{2 \alpha \phi}\delta(B_{np}) [H^{mnp}+H^{pmn}+H^{npm}] )=0 $$

as a total divergence term. The derivative acts on three terms, you forgot the exponential.

$$\partial_m [e^{2 \alpha \phi} (H^{mnp}+H^{pmn}+H^{npm})] = 0$$

Shouldn't be?

$$\nabla_m [e^{2 \alpha \phi} (H^{mnp}+H^{pmn}+H^{npm})] = 0$$